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Determine if $$\sum^{\infty}_{n=0} \dfrac{1+2^n}{3^n}$$ converges or diverges.

The tests I know so far are: by defintion, geometric, divergence, integral

Unfortunately none of them worked for me. Definition doesn't, this is not a geometric sum, divergence does not apply, and the integral is too complicated in this case.

What can I do?

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$$\sum_{n=0}^{\infty}\frac{1 + 2^n}{3^n} = \sum_{n=0}^{\infty}\frac{1}{3^n} + \sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^n.$$

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  • $\begingroup$ I got to this part but how do I evaluate the first series $\endgroup$ – K Split X Mar 23 '17 at 2:12
  • $\begingroup$ It's a geometric series so its equal to $$\frac{1}{1 - \tfrac{1}{3}}.$$ $\endgroup$ – Ethan Alwaise Mar 23 '17 at 2:12
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It converges since both $\sum (\frac{1}{3})^n$ and $\sum (\frac{2}{3})^n$ do.

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Hint:

  1. First solution: $$\frac{1+2^n}{3^n} = \frac{1}{3^n} + \frac{2^n}{3^n}$$
  2. Second solution: $$\frac{1+2^n}{3^n} \le 2\cdot\frac{2^n}{3^n}$$
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